2 Answers
2

To see that there's no bridges, assume that there is a bridge (say $uv$). Then as the graph is 4-regular, $u$ has four incident edges, and hence is itself incident to 3 faces (say $A$, $B$ and $C$), one of which is the "outer face" which is on both sides of the edge $uv$, but then we have a contradiction as all these faces are pairwise adjacent. So $A$ must have a different degree to $B$, without loss of generality assume these degrees are 3 and 4 respectively, but then $C$ is adjacent to $A$, so must have degree 4, but also adjacent to $B$ so must have degree 3.

The only wrinkle here is that the embedding does not actually need $uv$ to be on the outer face, but in this case we can see that it still is incident on only one face (otherwise it's not a bridge).

For the second part, let $e$ be the number of edges, $f_4$ the number of quadrilaterals, $f_3$ be the number of triangles and $v$ the number of vertices. Notice that by the regularity you get the number of edges in terms of vertices. In particular, you have $$4v=2e. $$ By a similar double counting argument: $$3 f_3+4 f_4= 2e.$$ Then you can use Euler's formula, which says $$f_3+f_4-e+v=2.$$
You also know something more, every triangle is adjacent to three quadrilaterals and by summing up the quadrilaterals this way, you overcount by a factor of $4$. Hence,
$$3 f_3 = 4 f_4.$$